Dirac Delta source for vectorial equation

[EmilyRuck]

Hello!
By manipulating Maxwell's equation, with the potential vector $\mathbf{A}$ and the Lorentz' gauge, one can obtain the following vector wave equation:

$∇^2 \mathbf{A}(\mathbf{r}) + k^2 \mathbf{A}(\mathbf{r}) = -\mu \mathbf{J}(\mathbf{r})$

The first step for the solution is to consider a point source $- \delta (\mathbf{r} - \mathbf{r}')$. But what about its meaning in a vector form?
A vector is a 1-dimensional object in the space; a point is a 0-dimensional object, according to linear algebra. How can we relate a 0-dimensional object to a vector context?
With such a source, the unknown function is a Green function and the previous equation becomes

$∇^2 G(\mathbf{r}, \mathbf{r}') + k^2 G(\mathbf{r}, \mathbf{r}') = - \delta (\mathbf{r} - \mathbf{r}')$

but now we don't have a vector function $\mathbf{A}(\mathbf{r})$ any more, but just a scalar function $G(\mathbf{r}, \mathbf{r}')$. Why?
This is like saying that, when the source is a point, the vector potential $\mathbf{A}(\mathbf{r})$ becomes scalar...?!
Thank you for having read,

Emily

 

[vanhees71]

The Helmholtz equation holds for each (Cartesian!) component of the vector field separately. So as soon as you have the Green's function for the scalar wave function you can apply it to each component.

BTW: Usually one needs the retarded Green's function, which gives you waves irradiated outwards from the sources. So you need to find a
$$G(\vec{r},\vec{r}') \propto \frac{\exp(+\mathrm{i}k|\vec{r}-\vec{r}'|)}{|\vec{r}-\vec{r}'|}.$$